In lecture my professor said that "weak convergence implies weak* convergence" but gave no explanation or proof, and ended class there. I'm trying to make sense of this statement but can't even see what is really being claimed here.
So first of all, what is the assumption of the claim? Clearly we have to be talking about a normed linear space $(X,\|\cdot\|)$. Are we assuming that there exists a weakly convergent sequence $\{f_n\}\rightharpoonup f$ in $X$? I feel like this must be the assumption, since any other quantifier seems silly.
Now what is the conclusion of the claim? That there exists a weak* convergent sequence? Or that $f_n$ determines a particular sequence $\varphi_n\in X^*$, and $f$ determines $\varphi$, such that $\varphi_n\overset w \to \varphi$ in $X^*$? I know that we can define a natural element in $X^{**}$ corresponding to any $g\in X$ by the definition $\psi_{g}(\phi') = \phi'(g)$, and this has the property that $\|\psi_{g}\|_{**}=\|g\|$ but it's not clear to me how this would determine some particular sequence of $\{\varphi_n\}\subseteq X^*$.
Best Answer
Your professor was just saying that if you have a sequence $(\varphi_n)$ in $X^*$ which converges weakly, then $(\varphi_n)$ also converges weak* (and both to the same limit). That is, weak convergence is stronger than weak* convergence.
Suppose that $\varphi_n\rightarrow\varphi$ weakly. Let $\Phi:X\rightarrow X^{**}$ be the identification map of $X$ and its double dual $X^{**},$ i.e. $\Phi(x)(\psi)=\psi(x).$ For any $x\in X,$
$$\varphi_n(x)=\Phi(x)(\varphi_n)\rightarrow \Phi(x)(\varphi)=\varphi(x),$$ using that $\varphi_n\rightarrow\varphi$ weakly. This shows that $\varphi_n\rightarrow\varphi$ weak*, as well.